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2 years ago

Can someone show me hwo to do this so that i know how to do this please

Mathematics

2 answers:

spayn [35]2 years ago

5 0

Y=y coordinate (2in this case)
x=x coordinate (-2 in this case)
m=slope (-2 in this case)
b= y int. (we dont know the number because it doesn’t label it so we just put the letter and you can solve for b if the question tells you so)

so the answer is
y=mx+b (slope int. form)
2=-2(-2)+b

spin [16.1K]2 years ago

3 0

Answer:

y = -2x - 2

Step-by-step explanation:

You might be interested in

The pointer of the vibration measuring instrument is observed to move between the 0.1 and0.3 marks on the vertical scale, when s

Norma-Jean [14]

Answer:

Step-by-step explanation:

Given points moves b/w 0.1 of 0.3 marks Hence amplitude is 0.2

Also , frequency (w) = 100 rad/s.

Natural frequency (wn) = √K/m

K = 20 N/mm x 1000mm/1m = 20000N/M

W = 1000mm/1m = 20,000N/M

FormulaX = FolK/1 - (w/wn)²

0.2 = b/20000/1 - (100/50)²

Given forcing frequency was doubled, W1 = 2 x 100=200rad/s

X1 =b/20,000/1 - (200/20)²

0.2/11 = 1 - (200/50)²/1 - (100/50)² = 0.2/11 = -15^5– 3

X1 = 0.04

8 0

3 years ago

The layer of gold coating on a ring should be 0.03 millimeters thick when the ring is completed. A coating of 0.021 millimeters

Arisa [49]

The amount of gold coating in millimeters needed for the ring is the difference between the total coating and the coating that has already been applied. From the given above, that would be,
0.03 mm - 0.021 mm = 0.009 mm
Thus, the answer to this question is letter B.

6 0

4 years ago

Read 2 more answers

Nikolay [14]

Answer:

Step-by-step explanation:

4 x 2= 8

3 x 5= 15

2(8 +15)= 2(23)=

23 x 2= 46

8 0

3 years ago

Read 2 more answers

The area of the trapezoid

ELEN [110]

Answer:

Step-by-step explanation:

The formula of an area of a trapezoid:

$A=\dfrac{b_1+b_2}{2}\cdot h$

b₁, b₂ - bases

h - height

We have b₁ = 10in, b₂ = 5in and h = 11in. Substitute:

$A=\dfrac{10+5}{2}\cdot11=\dfrac{15}{2}\cdot11=7.5\cdot11=82.5\ in^2$

7 0

4 years ago

Read 2 more answers

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

4 years ago